SGU 261. Discrete Roots （N次剩余）

N次剩余

题目：http://acm.sgu.ru/problem.php?

contest=0&problem=261

题意：给定n,a,p 求出x^n ≡ a(mod p)在模p意义下的全部解，当中p是素数

说明：

代码：

/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, n) for (int i = 0; i < n; i++)
#define debug puts("===============")
using namespace std;
typedef long long ll;

//高速幂
ll pow_mod(ll a, ll n, ll m) {
    ll res = 1;
    while(n) {
        if (n & 1) res = res * a % m;
        n >>= 1;
        a = a * a % m;
    }
    return res;
}

//求原根
vector<ll> a;
bool g_test(ll g, ll p) {
    for (ll i = 0; i < a.size(); i++)
        if (pow_mod(g, (p - 1) / a[i], p) == 1) return 0;
    return 1;
}
ll primitive_root(ll p) {
    a.clear();
    ll tmp = p - 1;
    for (ll i = 2; i <= tmp / i; i++) if (tmp % i == 0) { //这里还能够用筛素数优化
        a.push_back(i);
        while(tmp % i == 0) tmp /= i;
    }
    if (tmp != 1) a.push_back(tmp);
    ll g = 1;
    while(true) {
        if (g_test(g, p)) return g;
        g++;
    }
}

// 求离散对数
#define N 111111
struct node {
    ll x, id;
    bool operator < (const node & T) const {
        if (x == T.x) return id < T.id;
        return x < T.x;
    }
}E[N];
ll discrete_log(ll x, ll n, ll m) {
    int s = sqrt(m + 0.5);
    for (; (ll) s * s <= m; ) s++;
    ll cur = 1;
    node tmp;
    for (int i = 0; i < s; i++) {
        tmp.id = i, tmp.x = cur;
        E[i] = tmp;
        cur = cur * x % m;
    }
    sort(E, E + s);
    ll mul = pow_mod(cur, m - 2, m) % m; // mul = 1 / (x^s)
    cur = 1;
    for (int i = 0; i < s; i++) {
        ll more = (ll) n * cur % m;
        tmp.id = -1, tmp.x = more;
        int pos = lower_bound(E, E + s, tmp) - E;
        if (E[pos].x == more) return i * s + E[pos].id;
        cur = cur * mul % m;
    }
    return -1;
}

//扩展欧几里得
ll extend_gcd(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1, y = 0;
        return a;
    }
    else {
        ll r = extend_gcd(b, a % b, y, x);
        y -= x * (a / b);
        return r;
    }
}

//N次剩余
//给定n,a,p 求出x^n ≡ a(mod p)在模p意义下的全部解，当中p是素数
vector<ll> residue(ll p, ll n, ll a) {
    vector<ll> ret;
    if (a == 0) {
        ret.push_back(0);
        return ret;
    }
    ll g = primitive_root(p);
    ll m = discrete_log(g, a, p);
    if (m == -1) return ret;
    ll A = n, B = p - 1, C = m, x, y;
    ll d = extend_gcd(A, B, x, y);
    if (C % d != 0) return ret;
    x = x * (C / d) % B;
    ll delta = B / d;
    for (int i = 0; i < d; i++) {
        x = ((x + delta) % B + B) % B;
        ret.push_back(pow_mod(g, x, p));
    }
    sort(ret.begin(), ret.end());
    ret.erase(unique(ret.begin(), ret.end()), ret.end());
    return ret;
}
int main () {
    ll n, a, p;
    scanf("%lld%lld%lld", &p, &n, &a);
    vector<ll> ret = residue(p, n, a);
    printf("%d\n", ret.size());
    for (int i = 0; i < ret.size(); i++) printf("%d ", ret[i]);
    return 0;
}